Ask question

Ask question

All categories

2 years ago

5

A railroad car of mass 2,000 kg traveling at a velocity v = 10 m/s is stopped at the end of the tracks by a spring-damper system

, as shown in fig. 1. if the stiffness of the spring is k = 80 n/mm and the damping constant is c = 20 n-s/mm, determine:

1 answer:

Murrr4er [49]2 years ago

6 0

Stopped at the end of the tracks by a spg-damper system, as shown in fig. 1

You might be interested in

Which graph uses bars to show data that are broken into intervals?

AnnyKZ [126]

Answer:

A. scatter plot?

Explanation:

I dont really know if I'm right... sorry.

5 0

2 years ago

One end of a metal rod is in contact with a thermal reservoir at 745. K, and the other end is in contact with a thermal reservoi

Masteriza [31]

Answer:

a) $S_1=-9.65}\ J/K$

b) $S_2=71.18\ J/K$

c) 0 J/K

d)S= 61.53 J/K

Explanation:

Given that

T₁ = 745 K

T₂ = 101 K

Q= 7190 J

a)

The entropy change of reservoir 745 K

$S_1=-\dfrac{7190}{745}\ J/K$

Negative sign because heat is leaving.

$S_1=-9.65}\ J/K$

b)

The entropy change of reservoir 101 K

$S_2=\dfrac{7190}{101}\ J/K$

$S_2=71.18\ J/K$

c)

The entropy change of the rod will be zero.

d)

The entropy change of the system

S= S₁ + S₂

S = 71.18 - 9.65 J/K

S= 61.53 J/K

3 0

3 years ago

Gawaingnpang nkabuhayan,hamon at oportinidad

Savatey [412]

gawaingnpang nkabuhayan, hamon at oportinidad

7 0

2 years ago

Assoli18 [71]

Answer: 40.4M/s

Solution: 46.6/1.15 = 40.4347826 then round it to a single decimal point, since 3 is lower than 5 it will be rounded to 40.4

8 0

2 years ago

How much work must be done to bring three electrons from a great distance apart to 5.0×10^−10 m from one another (at the corners

Inessa05 [86]

Answer:

1.38 x 10^-18 J

Explanation:

q = - 1.6 x 10^-19 C

d = 5 x 10^-10 m

the potential energy of the system gives the value of work done

The formula for the potential energy is given by

$U =\frac{Kq_{1}q_{2}}{d}$

So, the total potential energy of teh system is

$U =\frac{Kq_{1}q_{2}}{d}+\frac{Kq_{2}q_{3}}{d}+\frac{Kq_{1}q_{3}}{d}$

As all the charges are same and the distance between the two charges is same so the total potential energy becomes

$U =3\times \frac{Kq^{2}}{d}$

K = 9 x 10^9 Nm^2/C^2

By substituting the values

$U =3\times \frac{9\times 10^{9}\times \ 1.6 \times 1.6 \times 10^{-38}}{5\times 10^{-10}}$

U = 1.38 x 10^-18 J

6 0

3 years ago